help with problem solving (1 Viewer)

[mathsbrain]

Trying to find help with the following 2 questions:
1)
Terry has a solid shape that has four triangular faces. Three of these faces are at right angles to each other, while the fourth face has side lengths 11, 20 and 21. What is the volume of the solid shape?
2) How many different ways can you put together 15 rectangle of sizes 1*2cm to form a 3*10cm rectangle?

Wondering if there's any textbook that can help improve on the "concept" that's involved in these kind of problem solving?

 

[Drdusk]

Wondering if there's any textbook that can help improve on the "concept" that's involved in these kind of problem solving?

There are books that teach this kind of problem solving. You want to look for books that are 'AMC' preparation books, AMC being the Australian Maths Competition. Questions are of the style your looking for so buying one of those books should give you what you're looking for.

 

[mathsbrain]

There are books that teach this kind of problem solving. You want to look for books that are 'AMC' preparation books, AMC being the Australian Maths Competition. Questions are of the style your looking for so buying one of those books should give you what you're looking for.

Thanks, but are those books only test papers or actually teaching problem solving strategies?
Anyways, I was wondering if you can help with the two problems given above?

 

[TheOnePheeph]

With the first one, you are given that 3 of the sides are perpendicular, so obviously the solid is a section of a rectangular prism. Now each face of these 3 perpendicular sides is a triangle, i.e. half of a rectangle, meaning you effectively have (1/2)*(1/3)=1/6of a rectangular prism. Also, each of these faces is obviously a right angled triangle The other triangular face has the side lengths as the hypotenuses of the 3 right angles triangles, and these are known. So anyway to find the volume of the full cube, you would find each unknown side length of these 3 perpendicular faces, as this effectively finds the dimensions. To do this, label each unknown side x, y, z, and use the pythagoras theorem relations on each known right angled triangle side to get a system of equations in x^2, y^2, z^2. Solving this you should get one side as 9, one as 2sqrt(10) and the other as 6sqrt(10). Multiplying these all together and dividing by 6, you will get the volume of the solid, which turns out to be 180.

Imo you are better at doing other types of maths for problem solving, the australian maths competition problems are all really silly convoluted problems.

 

[Drongoski]

I think the vol = (1/6) x 9 x 2sqrt(10) x 6sqrt(10) = 180 cubic units.

 

[TheOnePheeph]

I think the vol = (1/6) x 9 x 2sqrt(10) x 6sqrt(10) = 180 cubic units.

Yeah you are right, edited original post.

 

[mathsbrain]

thanks, how about the 2nd one? I have no clue where to begin

 

[HeroWise]

Damn i didnt see the 1/6th of rectangular prism x.x . Went straight for volumes by slices ahahha.

For b)
Recognise it is a rectangle, use combination approach to solve it. I am getting 756756 is that correct?

 

[TheOnePheeph]

Damn i didnt see the 1/6th of rectangular prism x.x . Went straight for volumes by slices ahahha.

For b)
Recognise it is a rectangle, use combination approach to solve it. I am getting 756756 is that correct?

This is not the answer I get, and I don't see how combinatorial techniques could help with this question, but feel free to explain how you've done it.

I have come up with a nice solution for b). Let f_n denote the number of ways of filling a 3×2n rectangle. First of all, consider the 3×0 rectangle. Obviously, this is empty and there is only 1 way of fitting it with rectangles, so f_0=1. Then consider the 3×2 rectangle. There are 3 ways of filling this, f_1=3. If you consider the 3×4 rectangle, first of all, consider the case in which it can be split up into a 3×2 rectangle and another 3×2 rectangle. To visualise this, let the rectangle be rotated so that the longest side is horizontal. Obviously, there will be 3×f_1 ways of filling this. Then, however, we need to consider the cases in which there exist no 2 separate 3×2 rectangles. For this to happen, the leftmost strip of the big rectangle must have 1 vertical rectangle (I will attach a diagram in a comment later to help visualise this). In order to allow the entire 3×4 rectangle to be filled up, we need to put another 2×1 rectangle 2 horizontal units away from this, i.e. in the rightmost position. Obviously, there are 2 positions for both of these rectangles, (bottom 2 rectangles vertically or top 2), but they must be in the same position, and once they are in place there is only one way to fill up the rectangle with 2×1s. This results in 2×f_0 for the case, meaning there is 3×f_1 + 2×f_0 =11 possible arrangements for a 3×4.

This general idea can then be extended. For a 3×6, simply split it into a 3×4 and a 3×2, to get 3×f_2 for the first case. Then consider the case where the vertical rectangle is 2 units away from the one at the far left, in which case you still have a 3×2 to fill, and 4 units away from the one on the far left. This results in 2×(f_1 + f_0), so the total is f_3=3×f_2 + 2×(f_1 + f_0). This can be easily generalised for f_n as f_n = 3×f_(n-1) + 2(f_(n-2)+.....+f_0). So to find f_5 for the question, simply find f_4 using the recurremce formula, then use it to find f_5. You should get 571.

Sorry for the confusing explanation, and there may be an easier way, but this is the way I saw to do it. I will try to attach a diagram

 

[TheOnePheeph]

Here is a diagram that will hopefully help with the explanation for b)

 

[mathsbrain]

Here is a diagram that will hopefully help with the explanation for b)

thanks for this but yeh I think theres a combinatorics way?

 

[TheOnePheeph]

thanks for this but yeh I think theres a combinatorics way?

Could be, I haven't thought about it too hard, but generally with questions like this a recurrence relation is the way to go. I haven't done anything like olympiad tutoring or much olympiad maths in general though so maybe there's a nicer way to do them with combinations.